Answer:
she used force to swing her ball all way back
Explanation:
To solve this problem we will use the concepts related to the resulting Vector Force product of two components, that is,
If we take the Force of 50 N as the force in the X direction and the Force of 40 N in the Y direction we will have to:
Finally, since Newton's second law, acceleration can be determined as
Therefore the resultant magnitude of the acceleration of the object is
The horizontal motion of a launched projectile affects its vertical motion is false
Answer: False
<u>Explanation:
</u>
As the trajectory falls on a plane, all the components like displacement, velocity can be classified in two components termed as horizontal and vertical component.
In a projectile motion, the horizontal components of displacement and velocity remains constant and there is change only in the vertical component of displacement and velocity.
Both components are not dependent to each other. So, the horizontal motions of a launched projectile does not affect its vertical motion.
As we know that when charge is released in electric field
It will have two forces on it
1. electrostatic force
2. gravitational force
now if the ball will accelerate upwards so we can say
net upward force = mass * acceleration
now we can find charge q on it by above equation
So above is the charge on the particle
Answer:
speed when it reaches y = 4.00cm is
v = 14.9 g.m/s
Explanation:
given
q₁=q₂ =2.00 ×10⁻⁶
distance along x = 3.00cm= 3×10⁻²
q₃= 4×10⁻⁶C
mass= 10×10 ⁻³g
distance along y = 4×10⁻²m
r₁ = = = 3.61cm = 0.036m
r₂ = = = 5cm = 0.05m
electric potential V =
change in potential ΔV =
ΔV = , where 2.00μC
ΔV =
ΔV = 2 × 9×10⁹ × 2×10⁻⁶ ×
ΔV= 2.789×10⁵
= ΔV × q₃
ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶
v² = 223.12 g.m/s
v = 14.9 g.m/s